Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 18 (a)/Solution 1

Following the hint, let's look for $A,B$ such that

$$f(x)={\frac {A}{2+x}}+{\frac {B}{1-x}}.$$

Following steps similar to those of Question 15, we find $A=B=2$. We must now notice that each summand on the right-hand side of

$$f(x)={\frac {2}{2+x}}+{\frac {2}{1-x}}$$

is the closed form of a geometric series $\sum _{n=0}^{\infty }ax^{n}$. Recall that this series converges to ${\frac {a}{1-x}}$, as long as $x\in (-1,1)$. Let's figure out how to express each summand as a series separately, since we are allowed to recombine them using Theorem 3.5.13 of [CLP].

For the first, we have

$${\frac {2}{2+x}}={\frac {1}{1+x/2}}={\frac {1}{1-(-x/2)}}=\sum _{n=0}^{\infty }\left({\frac {-x}{2}}\right)^{n}.$$

For the second, we have

$${\frac {2}{1-x}}=\sum _{n=0}^{\infty }2x^{n},$$

and we will analyse both radii of convergence in part (b) . It follows then that

$$f(x)=\sum _{n=0}^{\infty }\left({\frac {-x}{2}}\right)^{n}+\sum _{n=0}^{\infty }2x^{n}=\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2^{n}}}+2\right)x^{n}.$$

Therefore the sequence of coefficients is $A_{n}=\left({\frac {(-1)^{n}}{2^{n}}}+2\right)$.